JEE Main & Advanced AIEEE Solved Paper-2009

  • question_answer
    If the mean deviation of the numbers 1, 1 + d, 1 + 2d, ... , 1 + 100d from their mean is 255, then the d is equal to     AIEEE  Solved  Paper-2009

    A) 10.0

    B)                        20.0                       

    C)        10.1       

    D)        20.2

    Correct Answer: C

    Solution :

    \[\frac{|x-\overline{x}|}{n}=255\] \[\overline{x}=\frac{1+1+d+1+2d.....+1+100d}{101}\] \[=\frac{101+d\left( \frac{100\times 101}{2} \right)}{101}\] \[=1+\frac{50\times 101d}{101}=1+50\,d\] \[|x-\overline{x}|=225\times 101\] \[2d\left( \frac{50\times 51}{2} \right)=255\times 101\] \[d=\frac{225\times 101}{50\times 51}=10.1\]


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